Problem: Is the function given below continuous/differentiable at $x=1$ ? $g(x)=\begin{cases} x-1&,&x<1 \\\\ (x-1)^2&,&x\geq1 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Answer: Checking for continuity at $x=1$ For the function to be continuous at $x=1$, we need the two-sided limit $\lim_{x\to 1}g(x)$ to exist and be equal to $g(1)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 1^-}g(x)$ and $\lim_{x\to 1^+}g(x)$ exist and are equal to $g(1)$. According to $g$ 's definition, $g(1)=(1-1)^2=0$. $\lim_{x\to 1^-}g(x)$ $x-1$ evaluated at $x=1$ is equal to $0$. Since $x-1$ is continuous, we can be certain that $\lim_{x\to 1^-}g(x)=0$. $\lim_{x\to 1^+}g(x)$ $(x-1)^2$ evaluated at $x=1$ is equal to $0$. Since $(x-1)^2$ is continuous, we can be certain that $\lim_{x\to 1^+}g(x)=0$. We saw that the two one-sided limits exist and are equal to $g(1)$, so the function is continuous at $x=1$. Checking for differentiability at $x=1$ For the function to be differentiable at $x=1$, we need the two-sided limit $\lim_{x\to 1}\dfrac{g(x)-g(1)}{x-1}=\lim_{x\to 1}\dfrac{g(x)-0}{x-1}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 1^-}\dfrac{g(x)-0}{x-1}$ and $\lim_{x\to 1^+}\dfrac{g(x)-0}{x-1}$ exist and have the same value. $\lim_{x\to 1^-}\dfrac{g(x)-0}{x-1}=1$ $\lim_{x\to 1^+}\dfrac{g(x)-0}{x-1}=0$ The two limits exist, but they are not equal. Therefore, the function is not differentiable at $x=1$. Graphically, the function has a sharp turn at this point. [I would like to see that, please!] In conclusion, the function is continuous but not differentiable at $x=1$.